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Let:

$$g(x)=\frac{1}{1+e^{1/(x-1)}}$$

for $x\ne 1$ and $g(x)=a$ for $x=1$.

For what values of $a$ will $g(x)$ be continuous for every $x$?

Thanks in adavance!

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  • $\begingroup$ See here for a reference to make the math in your question much more readible. $\endgroup$
    – Daryl
    Dec 28 '12 at 12:21
  • $\begingroup$ Will apply this from now on. Thanks. $\endgroup$
    – pie
    Dec 28 '12 at 12:24
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Hint Look at $\lim\limits_{x\rightarrow1^+}g(x)$ and $\lim\limits_{x\rightarrow1^-}g(x)$ to determine what $a$ needs to be.

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    $\begingroup$ But since limg(x) from the right is different from the left, can the function ever be continues for all x? Even if a=1, then function still isn't continues. $\endgroup$
    – pie
    Dec 28 '12 at 12:28
  • $\begingroup$ @pie If the two one-sided limits are not equal, what can you say about the two-sided limit, and hence the continuity of the function at that point? $\endgroup$
    – Daryl
    Dec 28 '12 at 12:32
  • $\begingroup$ The two sided limit doesn't exist - but then the questions asks "For what values of $a$ will g(x) be continues for all x?". $\endgroup$
    – pie
    Dec 28 '12 at 12:34
  • $\begingroup$ @pie Looking at the graph here it is immediately obvious that no value of $a$ will make the function continuous at $x=1$. The limit analysis that you did confirms this conclusion. $\endgroup$
    – Daryl
    Dec 28 '12 at 12:38
  • $\begingroup$ Thank you, what a terrible wording that question has. $\endgroup$
    – pie
    Dec 28 '12 at 12:48

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